Tag: physics

Quizzes (598) Questions (13441)

Questions Related to physics

Multiple choice physics along with motion motion around us motion and rest moving things around us

 A person in a car tends to fall back when it suddenly starts. It is due to

  1. Inertia of rest

  2. Inertia of motion

  3. Inertia of direction

  4. none
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

when bus suddenly start moving in forward direction. This happens due to. When a bus suddenly starts, the standing passengers fall backward in the bus.

It is due to inertia of rest.
Hence,

option $A$ is correct answer.

Multiple choice physics along with motion motion around us motion and rest moving things around us

The wall of your classroom is in a state of -

  1. motion

  2. rest

  3. neither rest nor motion

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The walls are at rest with respect to the Earth. When you are sitting in the classroom, you are also at rest with respect to the Earth, and so the walls do not seem to be moving to you.

Multiple choice physics along with motion motion around us motion and rest moving things around us

If the forces acting on an object are balanced, then_____

  1. the object will be in equilibrium

  2. the object will be in motion

  3. the object will have zero acceleration

  4. the object loses its shape
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

according to newton's first law of motion, an object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

so if forces acting on the object is balanced then the object will be in equilibrium and it will not have any acceleration.
cannot comment about option B because if initially object is at rest then it will be always at rest .
so best possible answer are option A,C.

Multiple choice physics along with motion motion around us motion and rest moving things around us

If $\vec {u}=a \hat {i}+b \hat {j}+ c \hat {k}$ with $\hat {i},\hat {j},\hat {k}$ are in east, north and vertical directions, the maximum height of the projectile is ?

  1. $\dfrac{a^{2}}{2g}$

  2. $\dfrac{b^{2}}{2g}$

  3. $\dfrac{c^{2}}{2g}$

  4. $\dfrac{b^{2} c^{2}}{2g}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The maximum height of a projectile is H = (vertical component of velocity)²/(2g). If vector u = aî + bĵ + cĵ with î east, ĵ north (horizontal), and ĵ vertical, then the vertical component of initial velocity is c (assuming cĵ represents the vertical direction, which typically uses the unit vector k̂, but in this notation cĵ or ck̂ would give the vertical component c). Maximum height = c²/(2g).

Multiple choice energy in wave motion oscillation and waves waves physics

If the energy density and velocity of a wave are $u$ and $c$ respectively then the energy propagating per second per unit area will be

  1. $u/c$

  2. $c^2u$

  3. $uc$

  4. $c/u$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If the energy density and velocity of a wave are $u$ and $c$ respectively then the energy propagating per second per unit area will be $uc.$

Multiple choice energy in wave motion oscillation and waves waves physics

The kinetic energy per unit length for a wave on a string is the positional coordinate

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The kinetic energy for a particle is given by $(\mu \Delta x/ 2) (\dfrac{dy}{dt})^2$

Thus, it depends only on the time variable and not on the position variable

Multiple choice energy in wave motion oscillation and waves waves physics

A travelling wave has an equation of the form $A(x,t)=f(x+vt)$. The relation connecting positional derivative with time derivative of the function is:

  1. $\dfrac{dA}{dt}=\pm v^2 \dfrac {dA}{dx}$

  2. $\dfrac{dA}{dt}=\pm v \dfrac {dA}{dx}$

  3. $\dfrac{dA}{dt}=\pm \sqrt(v) \dfrac {dA}{dx}$

  4. $\dfrac{dA}{dt}=(2 \pi v/\lambda) \dfrac {dA}{dx}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Positional derivative and time derivative of a function f is $\dfrac{dA}{dt}=\pm v \dfrac {dA}{dx}$

The correct option is (b)

Multiple choice energy in wave motion oscillation and waves waves physics

Kinetic energy per unit length for a particle in a standing wave is zero at:

  1. nodes

  2. antinodes

  3. mid-way between a node and an antinode

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Particle at antinodes is momentarily at rest and hence has zero kinetic energy. Its speed comes down to zero at this point and all energy is stored in the form of potential energy.

Multiple choice energy in wave motion oscillation and waves waves physics

The total energy per unit length for a travelling wave in a string of mass density $\mu$ , whose wave function is $A(x,t) = f(x \pm vt)$ is given by: 

  1. $E _tot = \sqrt(\mu/2) (\dfrac{dA}{dt})^2$

  2. $E _tot = (\mu/2) (\dfrac{dA}{dt})^2$

  3. $E _tot = (\mu/2)^2 (\dfrac{dA}{dt})^2$

  4. $E _tot = (2\mu) (\dfrac{dA}{dt})^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a travelling wave y = f(x - vt), the energy density is related to the square of the partial derivative of the wave function with respect to time, specifically (mu/2) * (dy/dt)^2.

Multiple choice energy in wave motion oscillation and waves waves physics

The maximum potential energy / length increases with:

  1. Amplitude

  2. Wavelength

  3. Frequency

  4. Velocity
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If $y= A \sin (\omega t- kx)$ is the equation of a wave through a string, then the slope of the wave is $\dfrac{dy}{dx}=- Ak \cos (\omega t- kx)$. 

The maximum potential energy will be $T \times \Delta x \times (\dfrac{dy}{dx})^2A^2k^2=A^2k^2 \cos ^2(\omega t -  kx)T \Delta x$

The maximum potential energy will be obtained if cos (\omega t - kx)=1. Thus, maximum potential energy = $4 \pi^2A^2T f \times T \times \Delta x $; T is the tension in the string

We also know that $T=\mu v^2$. Substituting, we get, 

Maximum potential energy = $4 \pi^2 f^2A^2 \mu$
Thus maximum potential energy depends on frequency and as frequency increases, potential energy also increases

The correct option is (c)